Connect of exchange vector field .

Question

$D$ is Levi-Civita connect. $U,V,W,Z$ are vector field. $S$ is (0,4) tensor.

And $D^2_{X,Y}S=D_XD_YS-D_{D_XY}S$.

Are there any easy way to compute the below equation ?

If I unfold it, it is complex to compute ,so I hope there is some easy way.

Answer 1

Assume that $S$ is $(0,1)$-tensor (Other case are similar). If $E_i$ is coordinate vector field, first we will enumerate definitions and notations :

$$ D_k S_m:= (D_kS)_m $$

Here by definition of covariant derivative, $$ (D_kS)_m:= E_k (S_m) - S_n\Gamma_{km}^n $$

So

$$ D_iD_jS_k = E_i( D_jS_k)- D_jS_m \Gamma_{ik}^m $$

(Here $T:=D_jS$ is $(0,1)$-tensor That is above is $D_iT_k$)

$$ = E_iE_j(S_k)- E_i (S_m\Gamma_{jk}^m)- E_j(S_m) \Gamma_{ik}^m+ S_n\Gamma_{jm}^n \Gamma_{ik}^m $$

Hence by using symmetry of index $i, \ j$ and direct computation, $$ D_iD_jS_k - D_jD_i S_k = -S_m (E_i\Gamma_{jk}^m-E_j\Gamma_{ik}^m) + S_n( \Gamma_{jm}^n \Gamma_{ik}^m -\Gamma_{im}^n \Gamma_{jk}^m ) $$

$$ = S_m (-R_{ijk}^m) $$

And note that $ R_{ijk}^m=(D_i D_j E_k-D_jD_i E_k)^m $ Hence we complete the proof